Question

A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and height of the cone are 12 cm and 8 cm, respectively. Determine the ratio of the surface area of the hemisphere to the surface area of the cone.​

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: A \: toy \: is \: in \: the \: form \: of \: a \: cone \: surmounted \: on \: a \: hemisphere \\ so \: thus \: radius \: of \: base \: of \: cone = radius \: of \: base \: of \: hemisphere \\ so \: let \: the \: common \: radii \: be \: r$

$now \: for \: the \: conical \: part \\ diameter \: (d) = 12.cm \\ thus \: then \: radii \: (r) = 6.cm \\ height \: (h) = 8.cm$

$so \: now \: using \\ for \: conical \: part \\ l = \sqrt{r {}^{2} + h {}^{2} } \\ = \sqrt{(6) {}^{2} + (8) {}^{2} } \\ = \sqrt{36 + 64} \\ = \sqrt{100} \\ l = 10.cm$

$now \: we \: know \: that \\ surface \: area \: ie \: curved \: surface \: area\: of \: hemisphere = 2\pi.r {}^{2} \\ curved \: surface \: area \: of \: cone = \pi.rl$

$thus \: then \: \\ curved \: surface \: area \: of \: hemisphere \div curved \: surface \: area \: of \: cone = 2\pi.r {}^{2} \div \pi.rl \\ = 2r \div l \\ = (2 \times 6) \div 10 \\ = 12 \div 10 \\ = 6 \div 5 \\ = 6 :5$

$hence \: the \: ratio \: of \: the \: surface \: area \: of \: the \: hemisphere \: to \: the \: surface \: area \: of \: the \: cone. \: is \: 6:5$